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When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, thenA. work function of `A` is `2.25 eV`B. work function of `B` is `4.20 eV`C. `T_(A)=2.00 eV`D. `T_(B)=2.75 eV` |
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Answer» `p_(A)=(h)/(lambda_(A))`, `T_(A)=(P_(A)^(2))/(2m)=(h^(2))/(2ml_(A)^(2))`, `T_(B)=(h^(2))/(2mlambda_(B)^(2))=(h^(2))/(8mlambda_(A)^(2))` `(h^(2))/(2mlambda_(A)^(2))=4.25-phi_(A)`…..`(1)` `(h^(2))/(8mlambda_(A)^(2))=4.70-phi_(B)`…..`(2)` Also, `(h^(2))/(8mlambda_(A)^(2))=(h^(2))/(2mlambda_(A)^(2))-1.5` `(3)/(8)(h^(2))/(mlambda_(A)^(2))=1.5rArr(h^(2))/(mlambda_(A)^(2))=4eV` `:. T_(A)=2eV`, `T_(V)=0.5eV` `phi_(A)=2.25eV`, `phi_(B)=4.70-0.5=4.20eV` |
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